Question

Assume air resistance is negligible unless otherwise stated. Calculate the displacement in m and velocity in m/s at the following times for a rock thrown straight down with an initial velocity of 10.0 m/s from the Verrazano Narrows bridge in New York City. The roadway of this bridge is 70.0 m above the water. (Enter the magnitudes.)

Answer 1

Answer: 1) t=0.5 s; S=6.225 m and v=14.9 m/s

2) t=1 s; S=14.9 m and v=19.8 m/s

3) t=1.5 s; S=26.05 m and v=24.7 m/s

Step-by-step explanation:

The displacement
`S` is given by

`S=ut+(1)/(2) at^(2)`

and final velocity
`v` is given by

`v=u+at`

where
`u` is the initial velocity

`a` is acceleration

`t` is time taken

Case 1: when time is 0.5 s

The displacement is

`S=ut+(1)/(2) at^(2)\\S=10* 0.5 +(1)/(2)* 9.8* 0.5^(2)\\\\S=6.225 m`

the velocity is

`v=u+at\\v=10+9.8* 0.5\\v=14.9 m/s`

Case 2: when t=1 sec

The displacement is

`S=ut+(1)/(2) at^(2)\\S=10* 1 +(1)/(2)* 9.8* 1^(2)\\\\S=14.9 m`

the velocity is

`v=u+at\\v=10+9.8* 1\\v=19.8 m/s`

Case 3: t=1.5 s

The displacement is

`S=ut+(1)/(2) at^(2)\\S=10* 1.5 +(1)/(2)* 9.8* 1.5^(2)\\\\S=26.05 m`

the velocity is

`v=u+at\\v=10+9.8* 1.5\\v=24.7 m/s`